Can someone please explain in detail how you would go about with answering this question...I don't understand the answers in the back of the excel book.

Thank you so much!

The hard part in this one is really just that 1-marker.

Note that there is still the boring way of literally listing out every outcome (K,M), where K = what Kim spins and M = what Mel spins. Then just read off your table of 25 elements to see that 10 of them ensure Kim wins (and hence the probability is \(\frac25\)). In an exam, when in a panic, that would be a go-to option.

\[ \text{One way of going about it more cleverly is to use a concept}\\ \text{sometimes referred to as }\textbf{symmetric probabilities}. \]

\[ \text{There are 3 possible options for Kim, namely win, draw and lose.}\\ \text{Now note that for a draw}\\ \begin{align*}P(\text{Draw}) &= P(11)+ P(22)+P(33)+P(44)+ P(55) \\&= \frac1{25}+\frac1{25}+\frac1{25}+\frac1{25}+\frac1{25}\\ &= \frac15.\end{align*}\]

Alternatively, we note that the number of ways we can have a draw is \(5\) (because it's easy to see that a draw must occur when they both roll 1, 2, 3, 4 or 5). Then the probability of a particular outcome that results in a draw, say, 1 and 1, is \(P(11) = \frac15 \times \frac15 = \frac1{25}\). So therefore the probability of a draw is \(P(\text{Draw})=5\times \frac1{25} = \frac15\).

\[ \text{The symmetry now comes into play.}\\ \text{Observe that for the remaining configurations}\\ \text{there is a }\textbf{one-to-one}\text{ correspondence between Kim winning, and Kim losing.}\]

\[ \text{The one-to-one matching occurs because if we start with some outcome}\\ \text{where Kim wins, say Kim rolls 3 and Mel rolls 1,}\\ \text{we can immediately find an option where Kim loses by flipping their roles}\\ \text{namely Kim rolls 1 and Mel rolls 3 instead.}\]

\[ \text{Because we can do this for }\textbf{every}\text{ option Kim wins to one where she loses}\\ \text{we deduce that the number of different ways Kim can win}\\ \text{must }\textbf{equal}\text{ that of her losing.}\]

\[ \text{The consequence: }\boxed{P(\text{Win}) = P(\text{Lose})}.\]

Once we have this, we just note that since win, draw and lose are the only 3 outcomes, \( P(W) + P(D) + P(L) = 1\).

Therefore \( P(W) + \frac15 + P(W) = 1\), so \(P(W) = \frac25\).

The second part is now just a straightforward binomial probability question. If you haven't covered that in class you should come back to it later, but otherwise just sub into the formula.